Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in PYTHON3):

Solution:
import sys


def min_i_v(row):
    min_v = None
    min_i = None
    for i, v in enumerate(row):
        if min_v is None or v < min_v:
            min_v = v
            min_i = i
    return min_i, min_v


def dec_row(matrix, row_i, val):
    for i in range(len(matrix[row_i])):
        matrix[row_i][i] -= val


def dec_col(matrix, col_i, val):
    for i in range(len(matrix)):
        matrix[i][col_i] -= val


amount_of_rows, amount_of_cols = [int(a) for a in sys.stdin.readline().split()]
matrix = []
for i in range(amount_of_rows):
    matrix.append([int(a) for a in sys.stdin.readline().split()])

log = []
interrupted = False

if amount_of_cols >= amount_of_rows:
    for row_i, row in enumerate(matrix):
        min_i, min_v = min_i_v(row)
        dec_row(matrix, row_i, min_v)
        for i in range(min_v):
            log.append('row ' + str(row_i + 1) + '\n')
        col = [0] * amount_of_rows
        for c_i in range(amount_of_cols):
            for r_i in range(amount_of_rows):
                col[r_i] = matrix[r_i][c_i]
            min_i, min_v = min_i_v(col)
            if min_v < col[row_i]:
                interrupted = True
                break
            else:
                min_v = col[row_i]
                dec_col(matrix, c_i, min_v)
                for i in range(min_v):
                    log.append('col ' + str(c_i + 1) + '\n')
        if interrupted:
            break
else:
    col = [0] * amount_of_rows
    for col_i in range(amount_of_cols):
        for r_i in range(amount_of_rows):
            col[r_i] = matrix[r_i][col_i]
        min_i, min_v = min_i_v(col)
        dec_col(matrix, col_i, min_v)
        for i in range(min_v):
            log.append('col ' + str(col_i + 1) + '\n')
        for r_i, row in enumerate(matrix):
            min_i, min_v = min_i_v(row)
            if min_v < row[col_i]:
                interrupted = True
                break
            else:
                min_v = row[col_i]
                dec_row(matrix, r_i, min_v)
                for i in range(min_v):
                    log.append('row ' + str(r_i + 1) + '\n')
        if interrupted:
            break


if interrupted:
    sys.stdout.write('-1\n')
else:
    sys.stdout.write(str(len(log))+'\n')
    for i in log:
        sys.stdout.write(i)